PYQ JUNE 2021 SIFT 1

CSIR NET BIOLOGY (JUNE - 2021 Sift 1)
Previous Year Question Paper with Solution.

21. Following statements are made about uncompetitive inhibition of an enzyme:

A. Uncompetitive inhibitor binds to both free enzyme as well as an enzyme-substrate complex.

B. Addition of uncompetitive inhibitor lowers the Vmax of the reaction.

C. Apparent K_M of the enzyme is lowered.

D. Apparent K_M of the enzyme remains unchanged.

Which one of the following option represents the correct combination of the statments?

(a) B and C

(b) A and C

(d) A and D

Ans. (a)

Sol. In uncompetitive inhibition, an inhibitor binds at a site distinct from the substrate. However, an uncompetitive inhibitor will bind only to ES complex. In uncompetitive inhibition, apparent V_max and apparent K_m both decreases.

22. Following are the pKa's of the ionizable groups in lyine

pKa₁ = 2.16 (-carboxylic group)

pKa₂ = 9.06 (-carboxylic group)

pKa₃ = 10.54 (-carboxylic group)

Which one of the following option represents the pl of lysine?

(a) 7.25

(b) 5.61

(d) 9.8

Ans. (d)

Sol. pI = (pKa₂ + pKa₃)/2 = (9.06 + 10.54)/2 = 9.8.

23. The enzyme alkaline phosphatase was tested for it catalytic activity using the substrate para-nitrophenylphosphate. The K_M obtained was 10 mM and V_max was 100µmol/min. Which one of the following option represents the initial velocity of the reaction at a substrate concentrationof 10 mM?

(a) 50 µmol/min

(b) 100 µmol/min

(d) 20 µmol/min

Ans. (a)

Sol. Initial velocity, V₀ = 1/2V_max = 1/2 × 100 µmol/min = 50 µmol/min.

24. How many hydrogen bonds involving the backbone CO and NH can be observed in an -helix consisting of 15 amino acid residues?

(a) 10

(b) 11

(d) 13

Ans. (b)

Sol. 1 residue can be involved in maximum 2H-bonds, therefore 15 residues can make up to 2 × 15 = 30 H-bonds. In the -helix, 4 residues at the N-terminus and 4 at the C-terminus make only 1H-bond pre-residues. This makes the total number of H-bonds (30 – 2 × 4) = 22. When calculating this number, each H-bond is counted twice : one time for the donor residue and one time for the acceptor. The actual number of H-bonds is then 22/2 = 11.

25. Iron-sulphur clusters [Fe-S] are the key prosthetic groups that carry electrons in all of the below

EXCEPT:

(a) NADH - CoQ reductase

(b) Succinate – CoQ reductase

(d) CoQH₂ –Cytochrome C reductase

Ans. (c)

Sol. NADH – CoQ reductase – FMN and Fe-S.

Succinate – CoQ reductase – FAD and Fe-S.

Cytochrome C oxidase – Heme and Cu⁺.

CoQH₂ – Cytochrome C reductase – Heme and Fe-S.

26. A schematic of a metabolic pathway is shown below.

Under which of the followin conditions would stoichiometric amount of end products K and L be obtained if a concerted feedback inhibition mechanism were is operation.

(a) K inhibts F G and L inhibits F H; D E is inhibited at equal amounts of K and L

(b) D E is inhibited at equal amounts of K and L; K inhibits F H and L inhibitis F G

(d) K inhibits F H and L inhibits F G

Ans. (a)

Sol. Statement (a) is correct.

27. The following statements are being made of define the Mihaeli constant (K_M). It is :

A. Independent of enzyme concentration [E] and subtrate concentration [S].

B Equal to the dissociation constant when the [ES] complex dissociates more rapidly than product formaion.

C. Equal to dissociation constant when product formation is more rapid than [ES] complex dissociation.

D. An intrinsic property of an enzyme and does not depend on pH, temperature and ionic trength.

Which one of the following combination of statements is correct?

(a) A and B only

(b) A, B and D only

(d) A and D only

Ans. (a)

Sol. The lower value of K_m describes the greater affinity of the enzyme for the substrate. K_m is therefore, also a measure of the affinity of an enzyme for its substrate. The K_m is unique for each enzyme-substrate pair. Different enzymes act on the same substrate have different K_m values. The K_m value for an enzyme is independent of substrate and enzyme concentration but depends on the environmental conditions such as pH, temperature and ionic strength.

28. Progression across G1/S boundary followed by entry into S-phase is promoted by the activation of which one of the following protein complexes?

(a) Cdk4/Cyclin D

(b) Cdk2/Cyclin E

(d) Cdk4,6/Cyclin D, E

Ans. (b)

Sol. Cdk2/Cyclin E is require for the entry of cell from G₁/S to S phase of cell cycle.

29. In eukaryotic cells, covalently attached lipids help to anchor some water soluble proteins to the plasma membrane. One group of cytosolic proteins are anchored to the cytosolic face of membrane by a fatty acyl group (e.g. myristate or palmitate). These groups are generally covalently attached to which amino acids present at the N-terminus of the polypeptide chain?

(a) Glycine

(b) Tyrosine

(d) Lysine

Ans. (a)

Sol. A myristic acid (14-carbon, saturated fatty acid) molecule is attached to a protein through an amide linkage to the -amino group of an N-terminal Gly residue (myristoylation). A palmitic acid (16-carbon, saturated fatty acid) molecule is attached to a Cys residue close to the N or C-terminus via a thioester linkage (palmitoylation).

30. Progression across G1/S boundary followed by entry into S-phase is promoted by the activation of which one of the following protein complexes?

(a) Cdk4/Cyclin D

(b) Cdk2/Cyclin E

(d) Cdk4,6/Cyclin D, E

Ans. (c)

Sol. Importin proteins is required for transport of component (>50kDa) from cells cytoplasm to the nucleus. NLS (Nucleus Localizing Signal) containing cargo is bound by importin α and then importin B1 binds to importin to form a nuclear pore-targeting complex.

31. Following statements were made about chromosome cohesion during mitosis and meiosis.

A. Though cohesin is important for pairwise alignment of the chromosomes on the mitotic spindle, it is not important for the generation of tension across the centromere.

B. Cohesin binds to the chromosome even before the initiation of S-phase.

C. In fission yeast, centromere specific localization of Moa1 and Rec8 regulates the orientation of kinetochores at meiosis-I.

D. Cohesin exhibits uniform distribution/localization pattern across the chromosomal length.

E. Polo/Cdc5 is a positive regulator of Separase, an endopeptidase that facilitates opening of the cohesin ring.

Which one of the following combination contains all correct statments ?

(a) A, B and D only

(b) A, C and E only

(d) B, C and E only

Ans. (d)

Sol. Unlike in mitosis, in which sister chromatids are pulled apart; in anaphase I, homologous chromosomes are separated. In anaphase I, sister chromatids stay together. A special meiosis-specific cohesin termed Rec8 (expressed only during meiosis I) inhibits the separation of sister chromatids.

Cdc5 is a polo-like kinase required for mitotic exit; regulates mitotic spindle assembly. Enzyme separase, a cysteine protease, cleaves Scc1, a subunit of the cohesin which holds the sister chromatids together.

32. Given below are a few statements about nuclear transport.

A. RanGTP levels are higher in the nucleus than the cytoplasm.

B. Nuclear import receptors can shuttle between the nucleus and cytoplasm.

C. NTF2 transports RanGDP into the cytosol.

D. Export of mRNA is not directly dependent on Ran.

E. tRNA and miRNA export is mediated by exportins.

Which one of the following combination contains all correct statments ?

(a) A, B, C, D only

(b) B, C, D only

(d) A, C, E only

Ans. (c)

Sol. Nuclear transport factor 2 (NTF2) mediates the import of GDP-bound Ran (Ran-GDP) from the cytosol into the nucleus. The general nuclear export pathway for mRNAs does not rely directly on RanGTP-binding exportins. Instead, the process is driven by the ATP-dependent RNA helicase.

33. During cytokinesis, a small GTPase, RhoA, localizes to the equatorial membrane above the spindle midzone. The localization/activity of RhoA is potentially modulated by:

A. RhoGEF Ect2

B. Aurora B kinase

C. PLK1 kinase

D. MKLP1 kinesin

E. ATM and ATR

Which one of the following combination contains all correct statments ?

(a) A, B and D only

(b) A, B,C and D only

(d) A, C and D only

Ans. (b)

Sol. ATM (Ataxia Telangiectasia Mutated) and ATR (Ataxia Telangiectasia and Rad3 related) are closely related kinases that are activated by DNA damage. These serine-threonine protein kinases (members of the phosphatidylinositol 3'-kinase-like kinase family) do not participate in cytokinesis.

34. Cellular actin levels can be as high as 100-400 µM. Of this, unpolymerized actin concentration can be as much as 50-200 µM. However, the critical concentration for actin polymerization invitro is about 0.2 µM. Some of the following proteins inhibit polymerization of actin in cells.

A. Thymosin –

B. Capping protein CapZ

C. Tropomodulin

D. XMAP215

Which one of the following options lists all inhibitors?

(a) A, B and C only

(b) B, C and D only

(d) D, A and B only

Ans. (a)

Sol. -tubulin ring complex, a protein complex containing -tubulin and several accessory proteins, nucleates microtubule polymerization. However, -tubulin ring complex is not the sole microtubule nucleator. XMAP215 acts as microtubule nucleator in Xenopus egg extracts.

35. To delineate the steps in endoplasmic reticulum (ER) transport, a PhD student homogenized pancreatic acinar cells to isolate microsomes, which retain most of the biochemical properties of the ER. For this experiment, the student has planned a number of controls as mentioned below.

A. Treat one set of microsomes first with detergent and then with protease.

B. Treat one set of microsomes with protease only.

C. Treat one set of microsomes with micrococcal nuclease.

D. Treat one set of microsomes with detergent only.

Select the option that represents the best combination of the controls.

(a) A, B and D only

(b) B, C and D only

(d) B and D only

Ans. (a)

Sol. When cells are disrupted by homogenization, the ER breaks into fragments and reseals into small vesicles called microsomes. Micrococcal nuclease (MNase) is derived from Staphylococcus aureus and is relatively non-specific endo-exonuclease. Its substrates are both RNA and DNA.

36. To delineate the steps in endoplasmic reticulum (ER) transport, a PhD student homogenized pancreatic acinar cells to isolate microsomes, which retain most of the biochemical properties of the ER. For this experiment, the student has planned a number of controls as mentioned below.

A. Treat one set of microsomes first with detergent and then with protease.

B. Treat one set of microsomes with protease only.

C. Treat one set of microsomes with micrococcal nuclease.

D. Treat one set of microsomes with detergent only.

Select the option that represents the best combination of the controls.

(a) A, B and D only

(b) B, C and D only

(d) B and D only

Ans. (c)

Sol. There are two classes of transposable elements Class I (Retrotransposons) and Class II (Transposons or DNA transposons). Class I is of two types with LTR (e.g., Ty element in Drosophila, Copia in yeast) and without LTR retrotransposons. Non LTR retrotransposons are of two types : LINES (e.g., L1 element of human, R2 of Bombyx mori, Eye factor in Drosophila) and SINES (eg. Alu element of human and B1 element of micr).

37. In bacteria many of the tRNA genes do not contain the CCA sequence found at the 3' end of tRNA. In this context which one of the following statements represents the correct explanation?

(a) In these tRNAs amino acylation occurs at the 3' end of the tRNA irrespective of the presence of the - CCA sequence.

(b) CCA sequence is added to these tRNA transcripts in a DNA template independent manner.

(d) The absence of CCA sequence occurred only in the last common ancestor (LCA) during the course of evolution and the current day tRNA genes always possess a sequence to encode the CCA end of the tRNA.

Ans. (b)

Sol. A specialized RNA polymerase called a CCA-adding enzyme (or tRNA nucleotidyltransferase) adds sequence CCA post-transcriptionally. As its name indicates, this enzyme adds the terminal CCA to tRNAs that initially lack this sequence.

38. Some cells possess peptides which contain D- form of amino acids. How do they arise?

(a) These peptides are produced by ribosomes by incorporating D- amino acids at specific positions.

(b) Ribosome makes peptides with L-amino acids only. However, some of the amino acids in the peptides are replaced by D- amino acids by a pathway that involves excision of the L- amino acids.

(d) Peptides with D-amino acids exist only in archeae where they are made by the presence of racemases.

Ans. (c)

Sol. Peptides may be ribosomically and non-ribosomically synthesized. Non-ribosomically synthesized peptides (i.e., non-ribosomal peptides) are not specified by codons of mRNA and synthesis is not catalyzed by peptidyl transferase of ribosomes. The nature of amino acids in these peptides may be standard or non-standard and their absolute configuration may be L or D.

39. All of the following statements about bacterial transcription termination are true EXCEPT

(a) some terminator sequences require Rho protein for termination.

(b) inverted repeat and 'T' rich non- template strand define intrinsic terminators.

(d) Nus A is necessary for intrinsic transcription termination.

Ans. (d)

Sol. Nus A acts as an accessory protein and enhances intrinsic transcription termination.

40. Which one of the following proteins is essential for both the initiation of DNA replication as well as the continued advance of the replication fork?

(a) ORC

(b) Geminin

(d) Cdc6

Ans. (c)

Sol. The origin recognition complex (ORC) is a heterohexameric complex that specifically binds to origins of replication. During G1 phase of cell cycle, ATP-bound Cdc6 protein binds with ORC. Germinin is the inhibitor of cdt1 and it does not allow the binding of MCM protein. MCM helicase activation requires DDK and CDK along with some other proteins like cdc45 and GINS.

41. Which one of the following combinations represents the major protein or protein complex involved in chromatin condensation in yeast and human, respectively?

(a) HP1 and SIR Complex

(b) SIR complex and HP1

(d) SIR complex and Su(var)

Ans. (b)

Sol. The correct option is (d).

42. The basic difference between direct repair and base excision repair is

(a) Direct repair restores original structure of altered nucleotide without replacement, while in base excision repair the section of DNA containing the distortion is removed, the correct base is added and resealed.

(b) In direct repair, homologous recombination repairs the broken region while base excision repair restores original structure of altered nucleotide by modification.

(c) Direct repair restores original structure by non-homologous end joining without using homologous template while in base excision repair the section of DNA containing the distortion is repaired by using homologous recombination.

(d) In direct repair, an exonuclease, a DNA polymerase and a ligase are used, while in base excision repair a translesion polymerase that bypasses the bulky lesions is used by the cell.

Ans. (a)

Sol. Direct repair systems act directly on damaged nucleotides, converting each one back to its original structure. Base excision repair involves removal of a damaged nitrogenous base from nucleotide. cleavage of the sugar-phosphate backbone at the generated abasic AP site resynthesis with a DNA polymerase.

43. Which one of the following regions of the target gene is NOT used for making an RNAi construct to knock down its expression?

(a) 5' UTR of the mature transcript

(b) 3' UTR of the mature transcript

(d) Intronic region

Ans. (d)

Sol. The most important function of introns is that they allow for alternative splicing, making it possible to generate multiple proteins from a single gene. Some introns encode functional RNA molecules through further processing after they are spliced. Intronic region is not present in mature RNA so can't be used make RNAi construct.

44. Following statements were made about imprinting in the human genome.

A. Imprinting control centre (IC) harbors part of the SNRPN gene.

B. Imprinting of genes in an individual cannot be tissue specific.

C. Sperms and eggs exhibit identical pattern of genome methylation, except in the sex chromosomes.

D. At imprinted loci, expression depends on the parental origin.

Select the option with all the correct statements.

(a) A and D

(b) B and D

(d) B and C

Ans. (a)

Sol. Genomic imprinting (also called gametic or parental imprinting) is an epigenetic phenomenon that causes genes to be expressed in a parent-specific manner. Imprinting control regions (ICRs) are long-range cis-acting control sequences that confer imprinted monoallelic expression to several genes in imprinted gene clusters. The SNRPN gene is located on chromosome 15 and is imprinted and expressed from the paternal allele. It encodes a component of the snRNP complex, which functions in pre-mRNA processing and may contribute to tissue-specific alternative splicing.

45. Mitochondrial protein synthesis is of prokaryotic origin. Following statements are being made about the ribosomes from bacteria and mitochondria:

A. The bacterial ribosome consists of small and large subunits of 30S and 50S respectively, whereas in mitochondria of mammals these subunits are of 28S and 39S

B. In the bacterial ribosomes the RNA: protein ratio is about 2:1 whereas in mitochondria ibosomes this ratio is usually 1:2

C. Both the bacterial and mitochondrial ribosomes consist of 30S and 50S subunits

D. Both the bacterial and mitochondrial ribosomes consist of RNA and protein in the ratio of 1:1

Choose the option that represents all correct statments.

(a) A and B only

(b) B and C only

(d) A and D only

Ans. (a)

Sol. Mitochondrial protein synthesis is of Prokaryotic origin, the ribosome from bacteria and mitochondria have 30S/50S and 28S/39S as small and large subunit respectively.

In bacterial ribosome the RNA : Protein ratio is about 2 : 1 whereas in mitochondrial ribosome this ratio is 1 : 2.

46. Given below are a few statements about the . infection cycle

A. Competition between cI and cII gene products determines the establishment of lysogeny versus lysis.

B. cI binds O_R1 first while cro binds to O_R3 first

C. Cro binding to O_R represses cI transcription

D. Rich medium favours lytic cycle because cII is protected from cellular proteases

Which one of the following options represents all correct statements?

(a) B and C

(b) A and B

(d) A and C

Ans. (a)

Sol. Cro genes codes Cro protein, an inhibitor of CI synthesis and responsible for early regulation. CI repressor blocks the synthesis of Cro protein and also maintains its own concentration at a particular level. CI repressor binds with the three operators O_R1, O_R and O_R3. These operators have different affinities for the CI repressor, with O_R1 > O_R2 > O_R3. When there is too little CI repressor, only the O_R1 site, which blocks the cro gene expression, is filled.

47. Reproduction of × 174, a single stranded DNA phage involves several steps. A few statements are given below to explain the mechanism.

A. The single stranded × 174 DNA is converted into a double-stranded replicative form (RF).

B. Replication of double stranded replicative form results in the production of single stranded phages, about 50% of which are +ve sense phages and the remaining are –ve sense phages.

C. Replicationof the double stranded replicative from results in the production of only –ve sense phage.

D. Replication of the double stranded replicative form results in the production of only –ve sense phage.

Choose the option that correctly describe the process

(a) A only

(b) A and B

(d) A and D

Ans. (d)

Sol. The phi × 174 (or × 174) bacteriophage is a single-stranded DNA (ssDNA) virus that infects Escherichia coli, and the first DNA-based genome to be sequenced. Replication of double stranded replicative forms results in production of only -ve sense phage.

48. Which one of the following schematics depicts the potential relationship among the subunits IIo, IIa, and IIb of RNA polymerase II?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. RNA polymerase II (pol II), the enzyme responsible for all mRNA synthesis in eukaryotes, has been isolated in two forms, a 12-subunit "complete" enzyme and a 10-subunit "core." The two additional subunits of the complete enzyme, Rpb4 and Rpb7, form a heterodimer and associate reversibly with core. Schematic A is correct.

49. DNA replication occurs in S phase. The entry of cells into S phase is regulated by the tumor suppressor protein Rb. The statements below are made with reference to the role of Rb.

A. Rb binds to E2F in the cytosol and prevents E2F entry into the nucleus.

B. Rb is phosphorylated by cyclin A/ cdk 4.

C. Phospho Rb activates E2F.

D. E2F activates cyclin E production which promotes the G1/S transition.

Which one of the following options represents all correct statements ?

(a) A, B and C only

(b) B, C and D only

(d) C and D only

Ans. (c, d)

Sol. The active, un or hypophosphorylated form of the Rb protien binds transcription factor E2F and represses gene transcription which is required for transition from G1 to S phase. E2F acts as a transcription factor and induces many genes to transcribe. D-CDK4/6 and E-CDK2 phosphorylates Rb protein. Phosphorylation of Rb protein, reduces its affinity for E2F and allows E2F to heterodimerize with factor DP (DNA-binding proteins). E2F-DP complex induces the expression of critical late G1 genes, the products of which are required to prepare the cell for entrance into S-phase.

50. The following statements are made with reference to DNA replication:

A. Camptothecin causes intra- strand and inter- strand crosslinks in DNA, leading to stalling of replication forks.

B. Prevention of reinitiation of DNA replication during the same cell cycle is mediated by regulating the loading of the initiator complex ORC.

C. A glu ala mutation in the nucleotide building pocket of DNA polymerase III could lead to the incorporation of ribonucleotides in the extending DNA chain.

D. A mutation in the gene encoding Topoisomerase II could lead to entanglement of DNA daughter strands during replication.

Which one of the following options represents all correct statements ?

(a) A and B only

(b) B and C only

(d) B, C and D only

Ans. (c)

Sol. As the strands of DNA are separated at the replication fork due to DNA helicase, the dsDNA in front of the fork becomes increasingly positively supercoiled. If there were no mechanism to relieve the accumulation of these supercoils, the replication forks would halt. Enzyme topoisomerases are required to relieve the positive supercoiling that arise from DNA unwinding. Camptothecin is a topoisomerae inhibitor.

51. Which of the following is the causative agent of filariasis?

(a) Listeria monocytogenes

(b) Cryptococcus neoformans

(d) Brugiya malayi

Ans. (d)

Sol. The typical vector for Brugia malayi filariasis are mosquito species from the genera Mansonia and Aedes. During a blood meal, an infected mosquito introduces third-stage filarial larvae onto the skin of the human host, where they penetrate into the bite wound.

52. The cytoplasmic domain of the receptor of which of the following proteins does NOT function as tyrosine kinase?

(a) Epidermal growth factor

(b) Platelet derived growth factor

(d) Asialoglycoprotein

Ans. (d)

Sol. The major role of ASGPR is the binding, internalization, and subsequent clearance from the circulation of glycoproteins that contain terminal galactose or N-acetylgalactosamine residues.

RTK class I (EGF receptor family)

RTK class II (Insulin receptor family)

RTK class III (PDGF receptor family)

Cytosolic domain of all above function as tyrosine kinase.

53. Tumors are generally classified by

(a) the virus which caused them

(b) the person who discovered them

(d) the tissue or cell of origin

Ans. (d)

Sol. Both benign and malignant tumors are classified according to the type of cell from which they arise. Tumors are classified into four major groups according to their origin (epithelial, mesenchymal, hematopoietic and neuroectodermal). Each broad category has many subdivisions according to the specific cell type, the location in the body and the microscopic appearance of the tumor.

54. A breakthrough in cancer therapy is expected where T- cells are taken from a patient are modified in the laboratory to attack cancer cells before re-infusion in the patient. These T cells are called

(a) cancer associated receptor T cells

(b) chimeric antigen-receptor T cells

(d) cancer antigens recognition T cells

Ans. (b)

Sol. The special receptor is called a chimeric antigen receptor (CAR). Large numbers of the CAR T cells are grown in the laboratory and given to the patient by infusion. CAR T-cell therapy is used to treat certain blood cancers, and it is being studied in the treatment of other types of cancer.

55. The population that is at highest risk against influenza infection should be immunized annually. Which one of the following is the most important reason for this?

(a) Influenza virus can change its surface antigen very frequently.

(b) Influenza virus has a very short incubation period.

(c) Influenza virus has a reasonably longer incubation period giving memory B cells time to respond by producing high levels of serum antibody.

(d) Repeated immunization interferes with the differentiation of plasma cells from memory cells thereby decreasing levels of neutralizing antibody.

Ans. (b)

Sol. Children younger than 5 years old—especially those younger than 2—are at higher risk of developing serious influenza-related complications. An influenza vaccine is the best way to reduce the risk of getting sick with flu and developing any of the potentially that can result. Population at highest risk against influenza infection should be immunized annually because influenza virus has very short incubation period.

56. Localized increases in the cytosolic level of free Ca²⁺ are critical to its function as second messenger. Calmodulin, a small cytosolic protein, mediates many cellular effects of Ca²⁺. Which of the following is NOT CORRECT for Ca²⁺-calmodulin interaction?

(a) Each calmodulin molecule binds six Ca²⁺ ions in a cooperative fashion.

(b) Binding of Ca²⁺ causes calmodulin to undergo a conformational change leading to active calmodulin.

(c) Since binding of Ca²⁺ is cooperative, a small change in the level of cytosolic Ca²⁺ leads to a large change in the level of active calmodulin.

(d) One of the many enzymes activated by Ca²⁺- calmodulin is cAMP phosphodiesterase, which degrades cAMP and links Ca²⁺ and cAMP signaling.

Ans. (a)

Sol. Calmodulin (17 kDa) is a small, conserved, monomeric and calcium-binding acidic protein that is present in cytosol. It is present in all eukaryotes. One calmodulin binds with four calcium at a time.

57. Some cellular and extracellular proteins are enlisted in List I and their typical characteristics are enlisted in List II

Which one of the following is the most appropriate match ?

(a) A - I, B - II, C - III, D – IV

(b) A - II, B - III, C - IV, D – I

(d) A - II, B - IV, C - I, D - III

Ans. (c)

Sol. Nidogens are monomeric glycoproteins that contribute to a linkage of polymeric networks with stable affinity to the basal laminin and collagen IV. Fibronectin is a large, adhesive glycoprotein which is found in a number of locations, most notably on cell surfaces, in extracellular matrixes, and in blood. Integrins are heterodimers and function as transmembrane linkers between the extracellular matrix and the actin cytoskeleton. A cell can regulate the adhesive activity of its integrins from within. Vimentin one gene is present but many different isoforms are there due to alternative splicing.

58. Following statements have been proposed for cancer cells and cancer stem cells:

A. Cancer cells mostly have mutations whereas cancer stem cells do not.

B. Cancer cells divide to form two different populations of cells whereas cancer stem cells do not divide.

C. Cancer stem cells can undergo self-renewal whereas cancer cells cannot.

D. Cancer cells are predominantly resistant to chemotherapy and radiation.

E. Cancer stem cells are found only in the bone marrow and placenta.

Which one of the following combination of statements is correct ?

(a) A and C

(b) A and B

(d) C and D

Ans. (a)

Sol. Stem cells are characterized by the capacity for self-renewal and the ability to differentiate into diverse specialized cell types. This concept has been extended from the embryonic stem cells and adult stem cells to cancer stem cells and induced pluripotent stem cells. Cancer stem cells are a small subpopulation of cells within tumors with capabilities of self-renewal, differentiation, tumorigenicity and drug efflux.

59. Which of the following is/are associated with the presentation of endogenous antigens by Class I MHC molecule by an Antigen Presenting Cell (APC), given the condition that there is no crosspresentation of antigens by the APC?

(a) TAP1 and TAP2 proteins only

(b) Invariant chain (Ii)

(d) TAP1, TAP2 proteins and proteosome-like subunits LMP2, LMP7

Ans. (d)

Sol. The major histocompatibility (MHC) class I antigen presentation pathway plays an important role in alerting the immune system to virally infected cells. MHC class I molecules are expressed on the cell surface of all nucleated cells and present peptide fragments derived from intracellular proteins. LMP2 and LMP7 are proteins encoded by MHC genes that are tightly linked to the genes encoding TAP, the transporter that conveys peptides from the cytosol to the endoplasmic reticulum for assembly with MHC class I molecules. LMP2 and LMP7 are subunits of a subset of proteasomes, large molecular assemblies with multi-proteolytic activities believed to degrade damaged and unwanted cellular proteins.

60. Which one of the following statements with respect to development in amphibians is correct?

(a) Gastrulation begins with the invagination of bottle cells, followed by coordinated involution of the mesodermal precursors and the epiboly of the prospective ectoderm

(b) The organizer induces the Nieuwkoop centre

(d) In the absence of BMP inhibitors ectodermal cells form neural tube

Ans. (a)

Sol. The correct option is (a).

61. During normal development of sea urchin, .-catenin accumulates predominantly in the micromeres, which are fated to become endoderm and mesoderm. If GSK- 3 is blocked in the developing embryo:

(a) -catenin accumulation in the nuclei of large micromeres will be inhibited leading to formation of ectodermal ball.

(b) -catenin will accumulate in the nuclei of all blastula cells leading to an ectodermal ball.

(c) -catenin will accumulate in the nuclei of all blastula cells leading to animal cells getting specified as endoderm and mesoderm.

(d) -catenin which accumulate in the nuclei of large micromeres will be inhibited leading to animal cells getting specified as endoderm and mesoderm.

Ans. (c)

Sol. Accumulation of -catenin makes blastula cells to become endodermal and mesodermal. Blocking GSK-3, prevents phosphorylation of -catenin. Unphosphorylated -catenin accumulates and translocates to the nucleus, where it binds to LEF/TCF, displaces the co-repressor Groucho (Gro) and acts as a coactivator to stimulate the transcription of Wnt-responsive genes.

62. In which of the following stages of Arabidopsis embryogenesis do the visible distinctions between the adaxial and abaxial tissues of the cotyledons become initially apparent?

(a) Globular stage

(b) Zygotic stage

(d) Mature stage

Ans. (c)

Sol. In the torpedo stage of development, parts of the suspensor complex must be terminated. The suspensor complex is shortened because at this point in development most of the nutrition from the endosperm has been utilized, and there must be space for the mature embryo.

63. In a transplantation experiment, the area of presumptive ectoderm from an early frog gastrula was transplanted to a region of the newt gastrula destined to become parts of the mouth. The resulting salamander larvae had frog like mouth parts (frog tadpole suckers) instead of balancers as observed during development of wild type newt embryo. This is an example of

(a) Determination

(b) Genetic specificity of interaction

(d) Autonomous specification

Ans. (b)

Sol. Not all tissues can respond to the signal being produced by the inducer. For instance, if the optic vesicle (presumptive retina) of Xenopus laevis is placed in an ectopic location (i.e., in a different place from where it normally forms) underneath the head ectoderm, it will induce that ectoderm to form lens tissue. Only the optic vesicle appears to be able to do this; therefore, it is an inducer. However, if the optic vesicle is placed beneath ectoderm in the flank or abdomen of the same organism, that ectoderm will not be able to respond. Only the head ectoderm is competent to respond to the signals from the optic vesicle by producing a lens.

64. Which of the following 'R' gene products do not contain NBS-LRR domain?

(a) RPS2 protein of Arabidopsis

(b) Xa1 protein of rice

(d) Mlo protein of barley

Ans. (d)

Sol. NBS-LRR proteins are some of the largest proteins known in plants, ranging from about 860 to about 1,900 amino acids. They have at least four distinct domains joined by linker regions: a variable amino-terminal domain, the NBS domain, the LRR region, and variable carboxy-terminal domains. M1o protein of barley do not contain NBS-LRR domain.

65. What is the pattern of cleavage observed in mammals?

(a) Radial

(b) Spiral

(d) Bilateral

Ans. (c)

Sol. Radia – Echinodermata and Amphioxus

Spiral – Annelids, Molluscs and flatworms

Rotational – Mammals and Nematodes

Bilateral – Tunicates

66. Which one of the following statements is NOT correct regarding the tetrapod limb development?

(a) As the limb grows outward, the stylopod forms first, then the zeugopod and the autopod is formed last. Each phase is characterized by a specific pattern of Hox gene expression.

(b) The zone of polarizing activity (ZPA) is maintained by the interaction of the FGFs from the AER and Shh expressed from the mesenchyme.

(c) Although cell death in the limb is necessary for the formation of digits and joints, it is never mediated by the BMPs, which is only responsible for differentiating mesenchyme cells into cartilage.

(d) The dorsal-ventral axis is formed in part by the expression of Wnt7a in the dorsal portion of the limb ectoderm, which maintains expression level of Shh in the ZPA and Fgf4 in the posterior AER.

Ans. (c)

Sol. BMP activity also plays an indirect role in limb patterning as a key component of a feedback loop between SHH production in the posterior limb bud and fibroblast growth factor production in the overlying apical ectodermal ridge (AER).

67. The Dorsal protein is involved in generating the dorsal-ventral (DV) polarity in Drosophila. The following statements were made regarding the activity of the Dorsal protein in establishing the DV polarity.

A. In embryos that lack Gurken protein, the Dorsal protein is not translocated to the nucleus of the follicle cells which then causes ventralization of the embryo

B. Though Dorsal protein acts as a morphogen, it is found throughout the syncytial blastoderm of the early Drosophila embryo.

C. In embryos that lack Cactus protein the Dorsal protein can be found in the nucleus of cells with a ventral fate.

D. If the Dorsal protein is blocked from entering the nucleus, the genes responsible for specifying dorsal cell types are not transcribed.

Which of the above statements are correct ?

(a) A and B

(b) B and C

(d) A and C

Ans. (b)

Sol. The Gurken protein forms an anterior-posterior gradient along the dorsal surface of the oocyte. The Gurken signal is received by the follicle cells closest to the oocyte nucleus through a receptor protein, Torpedo. This signal causes follicle cells to differentiate into the more columnar dorsal follicle cells. Pipe synthesis is inhibited in dorsal follicle cells. Due to the short diffusibility of the signal, only the follicle cells closest to the oocyte nucleus (i.e., the dorsal follicle cells) receive the Gurken signal, which causes the follicle cells to take on a characteristic dorsal follicle morphology and inhibits the synthesis of pipe protein. Therefore, pipe protein is made only by the ventral follicle cells. Toll acts as receptor helps to activate protein Dorsal whose inhibitor is cactus. Active Dorsal enters the nucleus and provides ventral cell fate.

68. Following are certain statements regarding root growth and differentiation in plants:

A. Root hair, endodermis, xylem and phloem reach maturation in elongation zone of a developing root.

B. The root epidermal cells that are incapable of forming root hairs are called atrichoblasts.

C. Quiescent center is present just above root cap.

D. In Arabidopsis, an auxin transporter (ABCB4) plays a role in root hair emergence by maintaining intracellular auxin concentration.

Which one of the following combination of statements is correct ?

(a) A, B and C

(b) B, C and D

(d) A, B and D

Ans. (b)

Sol. Trichoblasts develop root hairs in the differentiation zone, while atrichoblasts become non-hair cells . QC's are to be found at the tips of growing roots, in the root meristem, and are typically surrounded by groups of initial cells.

69. Match the terms used in vertebrate limb development in List I with their descriptions in List II:

Which one of the following combination of the statements is correct?

(a) A - IV, B - III, C - II, D - V, E - I, F – VI

(b) A - I, B - II, C - III, D - IV, E - V, F – VI

(d) A - II, B - V, C - I, D - III, E - IV, F - VI

Ans. (a)

Sol. During embryonic development, sequential rounds of EMT and MET are needed for the final differentiation of specialized cell types and the formation of the three-dimensional structure of the organs. These sequential rounds are referred to as a primary EMT, secondary EMT and tertiary EMT . Mesenchyme directly gives rise to most of the body's connective tissues, from bones and cartilage to the lymphatic and circulatory systems. The Apical Ectodermal Ridge (AER) is one of the main signaling centers during limb development. It controls outgrowth and patterning in the proximo-distal axis. The progress zone is a layer of mesodermal cells immediately beneath the apical ectodermal ridge in the developing limb bud. The fate of the mesodermal cells is thought to be patterned by the length of time spent in the progress zone during limb outgrowth. The autopod, including the mesopodium and the acropodium, is the most distal part of the tetrapod limb, and developmental mechanisms of autopod formation serve as a model system of pattern formation during development.

70. Given below are some of the statements in connection with neural tube formation in vertebrates:

A. In primary neurulation the cells surrounding the neural plate direct the neural plate cells to proliferate, invaginate and separate from the surface ectoderm to form an underlying hollow tube.

B. In secondary neurulation the neural tube arises from the aggregation of mesenchyme cells into a solid cord that subsequently forms cavities to create a hollow tube

C. In birds primary neurulation generates the neural tube from anterior up to the hind limb developing region

D. In mammals, secondary neurulation begins at the level of sacral vertebrae

E. Anencephaly results when a failure to close the neural tube occurs, resulting in the forebrain remaining in contact with amniotic fluid.

Which one of the following options gives all correct statements?

(a) A, B, C, D and E

(b) A only

(d) C, D and E only

Ans. (a)

Sol. The neural folds migrate toward the midline of the embryo, eventually fusing to form the neural tube beneath the overlying ectoderm. The cells at the dorsalmost portion of the neural tube become the neural crest cells. Neurulation occurs in somewhat different ways in different regions of the body. During primary neurulation, the original ectoderm is divided into three sets of cells: (1) the internally positioned neural tube, which will form the brain and spinal cord, (2) the externally positioned epidermis of the skin, and (3) the neural crest cells. Secondary neurulation is a morphological process, it accounts for the formation of the caudal spinal cord in mammals including humans. A similar process takes place in birds.

71. Photochemically generated ATP is consumed in which one of the following phases of Calvin-Benson cycle?

(a) Only carboxylation

(b) Only regeneration

(d) Reduction and regeneration

Ans. (d)

Sol. The reduction stage of the C3 cycle utilizes ATP and NADPH to convert 3-PGA into glyceraldehyde-3-phosphate.

72. Dirigent proteins predominantly play an important role in biosynthesis of:

(a) lignans

(b) alkaloids

(d) amino acids

Ans. (a)

Sol. Dirigent proteins proteins are involved in plant defenses against pathogens and are proposed to mediate the free radical coupling of monolignol plant phenols to yield the cell wall polymers, lignans and/or lignin-like compounds.

73. Following are certain statements regarding tracheary elements of vascular plants

A. Xylem tracheids are highly elongated tapered cells that conduct water

B. Xylem vessel elements are less elongated and narrower than tracheids

C. Angiosperms may have both tracheids and vessel elements

D. Vessel elements are the only tracheary elements in almost all gymnosperms

Which one of the following options represents the combination of correct statements ?

(a) A, B and C only

(b) A and C only

(d) B and D only

Ans. (b)

Sol. Xylem is the principal water and minerals conducting tissue. It is composed of four different kinds of cells tracheids, vessel elements, xylem fibers and xylem parenchyma. The tracheids and vessel elements (together called trachery elements) are the conducting cells of the xylem. Tracheids are elongated and spindle shaped non-living cells. The walls of these cells are heavily lignified with openings in the walls called pits. Vessel elements (or vessel members) are short and cylindrical tube-like non-living cells and have perforated end walls. The part of the wall bearing the perforation or perforations is called the perforation plate. Gymnosperms lack vessels in their xylem.

74. Following are certain statements regarding phytochrome interacting factors (PIFs), a family of proteins that regulates photomorphogenic response in plants:

A. PIFs promote skotomorphogenesis by serving as transcriptional activators of dark induced genes.

B. PIFs on interaction with Pr get phosphorylated, followed by degradation via the proteasome complex.

C. The degradation of PIFs takes place in the presence of light.

D. PIF-induced genes are not expressed in light.

Which one of the following options represents the combination of correct statements ?

(a) A, B and C

(b) A, C and D

(d) B, C and D

Ans. (b)

Sol. Negatively acting transcription factors in PIFs (PIF1 to PIF8) are transcription factors that negatively regulate photomorphogenesis. The PIF proteins contain phytochrome domains in the N-terminal region and bHLH (basic helix-loop-helix) DNA binding and dimerization domains at their C-terminus. PIFs interact directly with the Pfr form of phytochromes and function as central repressors of phytochrome signaling.

75. The phytohormones ethylene (ET), methyl jasmonate (MeJA) and salicylic acid (SA) play important roles in plant defense. The following statements were made regarding induction of defensin PDF1.2 and pathogenesis related protein PR1:

A. ET/MeJA activates PDF1.2

B. ET/MeJA activates PR-1

C. SA activates PDF1.2

D. SA activates PR-1

Which one of the following options represents the combination of correct statements ?

(a) A and B

(b) A and D

(d) C and D

Ans. (b)

Sol. Activation of the plant defensin gene PIF 1.2 in Arabidopsis by pathogens has been shown previously to be blocked in the ethylene response mutant ein2-1 and the jasmonate response mutant coi1-1. Salicylic acid (SA) is a key signal for the activation of defense genes in response to stress. The activation of late defense genes by SA, such as PR-1, involves the participation of the NPR1 protein.

76. A researcher has obtained an Arabidopsis mutant defective in strigolactones (SLs), a novel plant hormone. The following statements were made regarding the mutant phenotype:

A. Shoot branching gets enhanced in the mutant plant

B. Hyphal branching of arbuscular mycorrhizal fungi (AM-fungi) gets enhanced during colonization in the mutant plants

C. Shoot branching gets inhibited in the mutant plants

D. Germination of seeds of parasitic plant is prevented near the mutant plant

Which one of the following options represents the combination of correct statements ?

(a) A and B

(b) B and C

(d) A and D

Ans. (d)

Sol. Strigolactones have ability to stimulate seeds germination of the parasitic plant Striga. They stimulate branching and growth of symbiotic arbuscular mycorrhizal fungi in the soil. They also inhibit axillary bud growth in plant and trigger the germination of parasitic plant seeds.

77. During light reaction in photosynthesis, electron is transported in electron transport chain (ETC) and produces ATP and NADPH in the process. Following are certain statements regarding ETC during light reaction:

A. Electron from P680 moves first to quinone and then to the pheophytin

B. P700 can receive electrons from plastocyanin

C. NADPH is produced at the end of light reaction

D. The hydrogen ions produced during light reaction gets concentrated in thylakoid lumen

Choose the correct answer from the options given below:

(a) A, B and C

(b) A, B and D

(d) B, C and D

Ans. (d)

Sol. During the light-dependent stage ("light" reactions), chlorophyll absorbs light energy, which excites some electrons in the pigment molecules to higher energy levels; these leave the chlorophyll and pass along a series of molecules, generating formation of NADPH (an enzyme) and high-energy ATP molecules. Statement B,C and D is correct.

78. Given below are various plant natural products and their basic structural unit:

Which of the following options represents the correct match of natural product and the basic unit:

(a) A - IV, B - I, C - III, D – II

(b) A - III, B - II, C - I, D – IV

(d) A - IV, B - III, C - I, D - II

Ans. (b)

Sol. Phenolics are aromatic benzene ring compounds with one or more hydroxyl groups produced by plants mainly for protection against stress. Alkaloids are a class of basic, naturally occurring organic compounds that contain at least one nitrogen atom. The terpenoids, also known as isoprenoids, are a large and diverse class of naturally occurring organic chemicals derived from the 5-carbon compound isoprene. Cyanogenic glycosides (CNglcs) are bioactive plant products derived from amino acids. Structurally, these specialized plant compounds are characterized as a-hydroxynitriles (cyanohydrins) that are stabilized by glucosylation.

79. Which one of the following is used in organification of tyrosine residues in thyroglobulin protein, during thyroid hormone biosynthesis?

(a) Iodine

(b) Reduced iodine

(d) Hydrogen iodide

Ans. (c)

Sol. The follicular cells of thyroid follicles actively accumulate iodide from the blood and secrete it into the colloid. In the colloid, iodide is oxidized into iodine (2I^– = I₂) and attached to a tyrosine residue of thyroglobulin protein. Thyroglobulin is also synthesized and secreted by follicular cells.

80. The dark current in retina is due to

(a) Closing of Na⁺ channels in the outer segment of photoreceptors

(b) Opening of K⁺ channel in the inner segment of photoreceptors

(d) Closing of K⁺ channels in the outer segment of photoreceptors

Ans. (c)

Sol. These channels are kept open by the presence of the second messenger cyclic guanosine monophosphate (cGMP), which in dark conditions is produced continuously by the enzyme guanylate cyclase. This phenomenon is known as the dark current.

81. Which one of the following is not present in the filtration slit diaphragm?

(a) NEPH1

(b) Paxillin

(d) NEPH2

Ans. (b)

Sol. Slit diaphragms, considered specialized adherens junctions, contain both unique membrane proteins (e.g., nephrin, podocin, and Neph1) and typical adherens junction proteins (e.g., P-cadherin, FAT, and catenins). Paxillin is a focal adhesion-associated protein that has a coordinative role in signaling pathways regulating cell shape, motility, and spreading.

82. The curve B in the figure below shows the oxygen dissociation profile at physiological concentration of CO₂ and at pH 7.

An increase in pH would lead to oxygen dissociation profile indicated by :

(a) curve B (no change in the dissociation profile)

(b) curve A

(d) curve D

Ans. (b)

Sol. Lowering of pH causes the oxygen affinity of the hemoglobin to decrease and shifts the curve to the right. An increase in pH causes the oxygen affinity to the hemoglobin to increase and shifts the curve to the left.

83. The mechanisms of thermogenesis in brown adipose tissue (BAT) in cold are described in the following proposed statements:

A. A thermogenic uncoupling protein, UCP1 helps in the heat production in BAT

B. Norepinephrine secretion from sympathetic nerve endings in BAT is decreased

C. Lipolysis is increased by low level of norepinephrine in BAT

D. A high content of mitochondria in BAT helps in the oxidation of fatty acids

E. Oxidation produces much heat as ATP synthase activity is low

Which one of the following options represents the combination of correct statements ?

(a) A, B and C

(b) A, D and E

(d) B, D and E

Ans. (b)

Sol. UCP1 is a BAT-specific protein located in the inner mitochondrial membrane, which uncouples the respiratory chain, producing heat instead of energy. This process is known as adaptive thermogenesis, and it is mediated by adrenergic stimulation. Fatty acid oxidation mainly occurs in mitochondria and involves a repeated sequence of reactions that result in the conversion of fatty acids to acetyl-CoA. As brown adipose tissue mitochondria contain very low amounts of ATP synthase, relative to respiratory chain components, they constitute a physiological system that allows for examination of the control of ATP synthase assembly.

84. The pressure in the 'space' between lungs and chest wall is known as intrapleural pressure. The following statements are related to the intrapleural pressure at different phases of respiration:

A. At the end of quiet expiration the tendency of the lung to recoil from chest wall is balanced by the recoil of chest wall in opposite direction, and intrapleural pressure is subatmospheric.

B. At the start of inspiration the intrapleural pressure is subatmospheric.

C. The intrapleural pressure becomes more negative during inspiration.

D. The intrapleural pressure attains value above atmospheric pressure during expiration.

E. The intrapleural pressure becomes positive (relative to atmospheric pressure) during strong inspiratory efforts.

Which one of the following combinations is correct ?

(a) A, B and C

(b) B, C and D

(d) A, C and D

Ans. (a)

Sol. During inspiration, the diaphragm and the inspiratory intercostal muscles actively contract, leading to the expansion of the thorax. The intrapleural pressure (which is usually -4 mmHg at rest) becomes more subatmospheric or more negative. During passive expiration, the diaphragm and inspiratory intercostal muscles cease contracting and relax, resulting in inward recoil of the chest wall and a decrease in the lung size. The intrapleural pressure increases to its baseline value, which decreases the TPP.

85. Hamoglobin (Hb) tranport CO₂ in venou blood as carbomates. The following statements refer to the formation of these carbomates:

A. CO₂ interacts with amino terminal nitrogens of Hb polypeptide chains

B. CO₂ interacts with carboxyl terminal carbons of Hb polypeptide chains

C. Carbonates helps formation of salt bridges between and chains of Hb

D. Carbonates helps formation of disulfide bridges between and chains of Hb

(a) A and C

(b) B and D

(d) A and D

Ans. (a)

Sol. CO₂ combines with the haemoglobin and forms carbaminohemoglobin. It combines with the amino groups of N-terminal amino acids of and -globin chains to form carbamates.

86. Following statements are given for the ovarian hormones:

A. 17 -estradiol, estrone and estriol are naturally occurring estrogens

B. They are 18 C steroids which do not have methyl group at 10th positions

C. They are 21 C steroids which have methyl group at 10th position

D. They are primarily synthesized by granulosa cells of the ovarian follicles

E. Their biosynthesis does not depend on the enzyme aromatase

Which one of the following options represents the combination of correct statements ?

(a) A, B and C

(b) A, B and D

(d) C, D and E

Ans. (b)

Sol. There are three major forms of physiological estrogens in females: estrone (E1), estradiol (E2, or 17-estradiol), and estriol (E3). Ovary steroid hormones are synthesized in both interstitial and follicular cells. The basic structure of cholesterol is three hexagonal carbon rings and a pentagonal carbon ring to which a side chain is attached. Two important methyl groups are also attached at positions 18 and 19.

87. Below are given a set of statements for the glucocorticoid hormones:

A. They bind to cell surface receptors and influence stress adaptation

B. They bind to intracellular receptors and influence stress adaptation

C. They inhibit ACTH secretion from anterior pituitary

D. Prolonged treatment with glucocorticoids leads to atrophic and unresponsive adrenals

E. Their secretion does not show circadian variations

Which one of the following combination of the statements is correct?

(a) A, C and E

(b) B, C and D

(d) A, D and E

Ans. (b)

Sol. The adrenal cortex produces steroid hormone glucocorticoids from zona fasciculata. Glucocorticoids are involved in carbohydrate metabolism. Glucocorticoids include cortisol (hydrocortisome), corticosterone and cortisone. The principal glucocorticoid is cortisol. The secretion of glucocorticoids is stimulated by ACTH from the anterior pituitary. ACTH promotes the secretion of cortisol.

88. DNA was isolated from a strain of bacterium with genotype a⁺ b⁺ c⁺ d⁺ e⁺ and transformed inot a bacteial strain a^– b^– c^– d^– e^–. The transformants were tested for the presence of the donated genes. The contransformed genes were found as follows :

a⁺ and b⁺; c⁺ and e⁺; d⁺ and c⁺;

What is the order of genes on the bacterial chromosome?

(a) a b c d e

(b) a c b e d

(d) a b d c e

Ans. (d)

Sol. The frequencies of cotransformation provide information about the relative distances between the genes. Genes located close to one another are more likely to be cotransformed; so, the frequency of cotransformation is inversely proportional to the distances between genes.

89. A fly with apricot coloured eye was crossed with a sepia eyed fly of opposite sex. In F1 all flies were wild type. The genes responsible for the two phenotypes were

(a) allelic

(b) non-allelic

(d) paralogous genes

Ans. (b)

Sol. The correct option is (b)

90. Human polydactyly traits having extra fingers or toes are caused by a dominant allele. In a screening it was found that out of 42 individuals having an allele for polydactyly, only 38 of them were polydactylus. Which of the following is the correct interpretation of the observation?

(a) The penetrance of polydactyly is estimated to be 90%

(b) The expressivity of polydactyly is 90%

(d) The polydactyly trait is showing complete penetrance

Ans. (a)

Sol. Penetrance of polydactyly = × 100 = 90.47% ~ 90%

91. The pedigree given below represents the genotype at four different loci for the children in genration III.

Which one of the given genotypes is likely to represent the genotype of individual II-1?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. C is the correct genotype of individual II-1.

92. A Drosophila mutant (line A) with vestigial wings is isolated in a laboratory. The vestigial wing phenotype was observed to be recessive and mapped to gene 'X'. Three other laboratories also isolated vestigial mutants, called line B, C and D. In order to test if the mutation in lines B-D also mapped to gene 'X', the following crosses were made and phenotype of F1 progeny observed.

Based on the above identify the line(s) which is most likely NOT to have a mutation in gene 'X'.

(a) Both lines B and C

(b) Line C only

(d) Both lines B and D

Ans. (c)

Sol. Line D only is known to have no mutation in gene 'X'.

93. 22 transduction was used to map the fadL gene. The result of two-factor crosses between fadL and two linked markers, purF and aroC are shown below.

Which one of the following is the correct map for the three genes?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Frequency of fadL purF⁺ = (200/1000) × 100 = 20%

Frequency of aroC⁺ fadL = (400/1000) × 100 = 40%

Frequency of aroC⁺ purF = (500/1000) × 100 = 50%

94. Wheat plants can have kernels of white colour or in shades of red i.e. light red, red, dark red and very dark red (purple).

A researcher made the following cross:

The following conclusions are made from the results obtained:

A. It is a dihybrid cross where white colour is coded by gene A and the purple colour is coded by gene B

B. Two genes, both coding for the colour of kernel and each gene having two alleles, one that produced red pigment and the other that produced no pigment.

C. Four genes, one coding for no pigment, which is epistatic over the other genes. The remaining three genes have 2 alleles each, one that produced red pigment and the other that produced no pigment.

D. The trait is influenced by the environment leading to the observed variation in kernel colour.

Which of the above conclusion(s) is/are correct?

(a) A only

(b) B only

(d) C and D only

Ans. (b)

Sol. The kernel color of wheat is a polygenic quantitative trait. Here more than one nonallelic genes determine the expression of a given trait. Hence, the relationship between genotype and phenotype is often very complex. In such traits, many different genotypes are possible, several of which may produce the same phenotype. Each non-allelic genes has two alleles one additive allele that contributes approximately equally to a specific phenotype and one non-additve allele that unable to contribute quantitatively to the phenotype. The contribution to the phenotype of each additive allele, though often small, is approximately equal. No allelic pairs exhibit dominance and there is no genetic interaction between nonalleles of different loci.

95. Eukaryotes are classified into 5-6 super groups based on phylogenomic studies. Which one of the following statements about eurkaryotic supergroups is FALSE?

(a) Fungi and animals are more closely related to each other than either group is to plants.

(b) Amoebozoa and opisthokonts are unikonts.

(d) Alveolates and amoeba belong to same super group.

Ans. (d)

Sol. Nearly all of eukaryotic diversity has been classified into 6 suprakingdom level groups (supergroups) based on molecular and morphological/cel-biological evidence; these are Opisthokonta, Amoebozoa, Archaeplastida, Rhizaria, Chromalveolata and Excavata.

96. Which of the following combinations would best characterize the dominant phase of the life cycle of a pteridophyte?

(a) Diploid gametophyte

(b) Haploid gametophyte

(d) Haploid sporophyte

Ans. (c)

Sol. The diploid zygote is the first stage in the sporophyte generation. The zygote divides by mitosis and develops into a multicellular embryo, the young sporophyte plant. In case of land plants when we move from bryophytes to angiosperms, dominance of increases.

97. Which one of the following gases is present in the stratosphere at a concentration higher than its concentration in troposphere?

(a) Nitrogen

(b) Oxygen

(d) Carbon dioxide

Ans. (c)

Sol. A high concentration of ozone, a molecule composed of three atoms of oxygen, makes up the ozone layer of the stratosphere.

Ozone an unusual type of oxygen molecule that is relatively abundant in the stratosphere, heats this layer as it absorbs energy from incoming ultraviolet radiation from the sun.

98. The graph below shows the change in size four population (A-D) over time.

Which among the four populations (A, B, C and D) would have the lowest intrinsic rate of population growth (r)?

(a) A

(b) B

(d) D

Ans. (d)

Sol. The intrinsic rate of increase or per capita rate of increase can be maximum or realized. The maximum per capita rate of increae (r_max) occurs during exponential population growth under an ideal unlimited environment. The realized or actual per capita rate of increae (r) occurs under an actual prevailing environment.

99. Consider the following ecosystems.

A. Tropical rain forests

B. Open ocean

C. Algal beds and Coral reefs

D. Marshes and Swamps

Which one of the following options represents these ecosystems in an increasing order of their contribution to annual world net primary production?

(a) B, C, D and A

(b) C, D, B and A

(d) C, D, A and B

Ans. (d)

Sol.

100. Historical frequencies of fires in an area can be determined by

(a) radioactive dating of the tree remains.

(b) measuring carbon content on the soil surface after fire.

(d) examining records of evacuation history of the nearby villages.

Ans. (c)

Sol. Methods to describe past fire activity include the use of historical documents, fire-scar tree rings, and forest stand age measurements over the past few centuries.

101. Given below are two columns listing angiosperm families and their groups

(a) A - IV, B - II, C - I, D – III

(b) A - IV, B - I, C - III, D – II

(d) A - II, B - III, C - I, D - IV

Ans. (a)

Sol. A. Basal angiosperms : Nympheaceae

B. Fabids : Cucurbitaceae

C. Malvids : Brassicaceae

D. Lamids : Solanaceae

102. Given below are the names of diseases caused in rice in Column X and the names of the diseasecausing organisms in Column Y.

Which one of the following options represents the match of column X and Y ?

(a) A - II, B - I, C - V, D - IV, E – III

(b) A - III, B - IV, C - II, D - V, E – I

(d) A - III, B - V, C - II, D - I, E - IV

Ans. (d)

Sol. A. Bacterial blight : Xanthomonas oryze pv. oryzea

B. Garin rot : Burkholderia glumac

C. Sheath blight : Rhizoctonia solani

D. Leaf smut : Entyloma oryzae

E. Downy mildew : Sclerophthora macrospora

103. The table given below lists the morphological features and groups of plants.

Which one of the followings options represents the correct match between the two columns?

(a) A - I, III and V ; B - II, III, and V

(b) A - I, III and IV ; B - II and IV

(d) A - I and V ; B - II, IV, and V

Ans. (d)

Sol. Liverworts have unicellular rhizoids that are located on the ventral (bottom) side of the gametophyte. Rhizoids look like roots, but do not absorb water or nutrients. Instead, they attach the plants to their substrate and help with external water retention and conduction. Mosses are flowerless small plants found under the division Bryophyta along with liverworts and hornworts. They do not possess any vascular system like xylem and phloem, and mainly absorb water and nutrients through their leaves. They are mostly found in damp, shady locations as mats or clumps on the forest floor.

104. The table given below lists fossils and the major group of plants to which they belong.

Whcih one of the following options represents the correct match between columns ?

(a) A - I, B -III, C - IV, D – II

(b) A - III, B -II, C - IV, D – I

(d) A - II, B -III, C - I, D - IV

Ans. (b)

Sol. A. Naiadita lanceolata : Bryophyte

B. Rhynia gwyne-vaughanii : Pteridophyte

C. Antarticycals schopfii : Bryophyte

D. Tricolpites minutes : : Angiosperm

105. The table given below lists the conservation areas and the major group of organisms that they are best known for.

(a) P-1, Q-2, R-3, S-4, T-5

(b) P-5, Q-2, R-3, S-4, T-1

(d) P-5, Q-2, R-4, S-3, T-1

Ans. (b)

Sol. Hemis National Park is home to some of the most exotic and rare flora and fauna species. The most sought after animal species of the sanctuary are Snow Leopards, Shapu, Wild Sheeps, Ibex and Goats.

National Chambal Sanctuary, also called the National Chambal Gharial Wildlife Sanctuary, is a 5,400 sq. km tri-state protected area in northern India home to critically endangered gharial (small crocodiles), the red-crowned roof turtle and the endangered Ganges river dolphin.

Nokrek national park - endemic citrus species.

Sessa sanctuary - Orchids

Baghmara wildlife sanctuary - Nepenthes.

106. The table given below lists types of extremophiles and the environments that they are adapted to.

Which one of the following options represents the correct match of the columns?

(a) A - I, B - II, C - III, D - IV, E – V

(b) A - V, B - II, C - III, D - IV, E – I

(d) A - V, B - II, C - IV, D - III, E - I

Ans. (c)

Sol. A. Psychrophile : Extreme cold temperatures

B. Hyperthermophile : Extreme high temperature

C. Alkaliphile : High alkaline environment

D. Hyperpiezophile : High pressure

E. Halophile : High salinity

107. The table below show the bird species and their abundance in three habitats P, Q and R.

Which one of the combinations below best represents the habitats in decresing order of diversity?

(a) P, R, Q

(b) R, Q, P

(d) Q, R, P

Ans. (d)

Sol. The number of species and their relative abundance define species diversity of a community. It includes the number of species in a community (i.e., richness) and the relative abundance of each species (i.e., evenness). Species richness is simply the number of species in a community. But, among the array of species that make up the community, not all are equally abundant.

When every species in a sample has the same abundance, species evenness is at its maximum level. Habitat P has dominance based abundance, so it will be less diverse. Habitat Q has almost similar abundance, so it wil be more diverse.

108. The following are a few statements on shade leaves vis-à-vis sun leaves in tree species

A. Higher amount of chlorophyll per dry weight

B. Lower density of stomata

C. Thicker leaves

D. Lower rates of dark respiration per unit area

Which one of the following combinations of above statements is correct?

(a) A and D

(b) B and C

(d) B and D

Ans. (c)

Sol. Shade leaves has to perform efficient photosynthesis. Therefore, it has higher amount of chlorophyll per dry weight. Shade leaves are darker in colour because the chloroplast do not mask each other. It has low rate of transcription hence low density of stomata. To minimize carbon loss, dark leaves has lower rates of dark respiration per unit area.

109. The diagram below depicts energy flow within a single trophic level, where I = amount ingested, NA = amount not asimilated, R = respiration and Pn = biomass production at trophich level.

Which one of the following options represent correct value for Pn, NA, R and I in kcal respectively, if Pn–1 = 1000 kcal, I/Pn–1 = 20%, A/I = 35% and Pn/A = 20%.

(a) 56 14 130 200

(b) 14 130 56 200

(d) 56 130 200 14

Ans. (b)

Sol. Consumed biomass = I = = 200

Assimilated amount = A = = 700

NA = amount not assimilated = 200 – 70 = 130

Biomass production = (P_n) = = 14

Respiration = R = 70 – 14 = 56.

110. The following graph shows the change in proportion of biomas in foliage (leaves), branch and stemwood (bole) for a trees species as a functino of DBH (diameter at 1.5 m above ground)

Which one of the following options correctly matches the curves (A, B and C) with stemwood, foliage and branch respectively?

(a) A, B and C

(b) A, C and B

(d) B, A and C

Ans. (b)

Sol. Biomass growth continuously increases with tree size for cross-sectional analyses. Longitudinal analyses are less likely to show continuously increasing biomass growth with size. Increasing biomass growth rates are correlated with increasing rates of light absorption.

111. The graph given below show the possible behaviour of two species over the course of succession.

Possible effects observed during succession are:

i. Total suppression

ii. Convergene

iii. Sequential succession

Choose the correct option matching the graphs with the possible effect :

(a) A-(i) B-(ii) C-(iii)

(b) A-(ii) B-(iii) C-(i)

(d) A-(iii) B-(i) C-(ii)

Ans. (c)

Sol. The correct option is (c).

112. Consider the following graph for per capita growth rate as a function of population density (N).

Which one of the plots correctly depicts strong Allee effect in a population?

(a) A

(b) B

(d) D

Ans. (c)

Sol. Density-dependent factors can have either a positive or a negative correlation to population size. As population size increases, either birth rate declines or mortality rate increases or both. It is a negative feedback. However, not always density-dependent factors are negatively related to population size. In some cases, growth rate increases with population size. This phenomenon is referred to as the Allee effect (after W. Allee, who first described it) and is an example of positive feedback.

113. In the Table below, Column I describes movements of organisms and Column II describes the type of movement.

Which one of the following options represents the correct match of column I with Column II

(a) A - IV, B - III, C - I, D – II

(b) A - IV, B - III, C - II, D – I

(d) A - I, B - II, C - III, D - IV

Ans. (a)

Sol. Mnemotaxis : a taxis involving a constant reaction (such as movement at a constant angle to a light source) but not a simple movement toward or away from the directing stimulus.

Mnemotaxis, literally "memory movement", describes navigation through the use of landmarks. Many birds navigate using landmarks, as do salmon. Humans also use mnemotaxis when navigating through the use of street signs and familiar buildings.

Magnetotaxis is a process implemented by a diverse group of Gram-negative bacteria that involves orienting and coordinating movement in response to Earth's magnetic field.

Klinotaxis is the achievement of orientation by alternate lateral movements of part or all of a body; there appears to occur a comparison of intensities of stimulation between one position and another and a "choice" between them.

114. A researcher working on island biogeography mapped how isolation controlled immigration (I), and area controlled extinction (E), will act on number of species present on the island. He forgot to label the size of the island (small or large) and the locatio of the islands (near or far) on the graph.

Using information from MacaArthur and Wilson's equilibrium theory, select the option that correctly identifie A, B and C and D in the figure above.

(a) A-large, B- small, C-near, D-far

(b) A-small, B-large, C-far, D-near

(d) A-far, B-near, C-large, D-small

Ans. (c)

Sol. Effect of island size : Large island have a larger equilibrium number of species than small islands because extinction rates lower on large islands. Effect of distance from mainland : Near islands tend to have larger equilibrium numbers of species than far islands because immigration rates to near islands are higher. If the island is close to the mainland, it will be easy for species to reach, so the emigration rate will be very high. If the island far from mainland, then fewer species will reach it and the rate of immigration will be much lower.

115. Given below are statements on 'living fossils'. Select the correct statements.

(a) Living fossils are impressions of extant organisms in old rocks.

(b) Living fossils show high morphological divergence from fossil records.

(d) Living fossils are organisms that have remained unchanged for millions of years.

Ans. (d)

Sol. A living fossil is an extant taxon that cosmetically resembles related species known only from the fossil record. To be considered a living fossil, the fossil species must be old relative to the time of origin of the extant clade.

116. Mimicry where deceptiveness of the mimic's signal is high and fitness consequences signaled to the receiver by the mimic is also high (and negative) is

(a) Batesian mimicry

(b) Müllerian mimicry

(d) Millerian mimicry

Ans. (a)

Sol. In Batesian mimicry, a vulnerable organism (termed mimic) gains a selective advantage by looking like a dangerous of distasteful organism (termed model). The classic example of Batesian mimicry are the monarch (Danaus plexippus) and viceroy (Limenitis archippus) butterflies.

117. Consider the following four geological periods.

A. Quaternary

B. Cretaceous

C. Jurassic

D. Cambrian

Which one of the following options represents the correct arrangement of these geological periods from earliest to recent:

(a) A-B-D-C

(b) D-C-B-A

(d) B-A-C-D

Ans. (b)

Sol. The correct option is (b).

118. Select the correct statement related to phylogeny of primates.

(a) Lemurs are more closely related to lorises than to gibbons.

(b) Orangutans are closer to lorises than to gibbons.

(d) Humans are closer to new world monkeys than to orangutans.

Ans. (a)

Sol. Phylogeny of primates

119. Which one of the following is valid with respect to, one step growth experiment developed by Ellis and Delbruck in 1939?

(a) The reproduction of large phage population is synchronized.

(b) A culture is directly developed by inoculation of single bacterial colony from the agar plate into liquid medium.

(c) Involves only a single step of overnight culture development followed by inoculation of a fresh medium with 1% inoculum.

(d) Only a single carbon source such as glucose is used in the medium.

Ans. (a)

Sol. First performed by Ellis and Delbruck in 1939, this classic experiment illustrates the true nature of virus replication. The "One Step Growth" experiment allows the calculation of both the duration of the different phases (see above) and the yield of the viral cycle (also known as burst size, that is, the number of virions produced by each infected cell).

120. In a population that is in Hardy-Weinberg equilibrium, the frequency of the recessive homozygote genotype of trait q is 0.04. The percentage of individuals homozygous for the dominant allele is

(a) 64

(b) 40

(d) 16

Ans. (a)

Sol. Given that, q² = 0.04, So, q = 0.2

p = 1 – q = 1 – 0.2 = 0.8

p² = 0.8 × 0.8 = 0.64

Percentage of individuals homozygous for the dominant allele = 0.64 × 100 = 64%

121. A researcher is interested in investigation if parental investment (PI, panel A) by male seahores and pipefishes is correlated with their mating patterns (MP-monogamy and polygamy, panel B). For this, the researcher build a phylogenetic tree of seahorses and pipefishes and maps PI and MP score on the tree as shown in figure below.

Based on the trees generated, which one of the following conclusions can be researcher correctly arrive at?

(a) Polygamy is correlated with simpler brood pouches.

(b) Monogamy is not correlated with elaborate brood pouches.

(d) Polygamy is correlated with elaborate brood pouches.

Ans. (b)

Sol. Many species appear to be monogamous (a rarity in nature), and all male syngnathids have some form of specialized egg-brooding structure located on the abdomen or tail, into which females deposit their eggs during mating. The males' brood pouches vary in complexity among species, from simple sticky patches to complex uterus-like structures with placenta-like features for nourishing the eggs: a mere transvestite seems unadventurous by comparison.

122. Given below are graph depicting two possible of gene duplication events over a period of time during genome evolution.

Based on the above figures, which one of the following options correctly represents the identify of A, B, C and D?

(a) A-Gene duplication event, B-random loss of duplicated genes, C-remaining pairs of duplicated genes, D-additional gene duplication events

(b) A-remaining pairs of duplicated genes, B-gene duplication event, C-random loss of duplicated genes, D-additional gene duplication events

(c) A-additional gene duplication events, B-random loss of duplicated genes, C-remaining pairs of duplicated genes, D- Gene duplication event

(d) A-Random loss of duplicated genes, B-additional gene duplication events, C-gene duplication event, D-remaining pairs of duplicated genes

Ans. (a)

Sol. Gene duplication is an important mechanism for acquiring new genes and creating genetic novelty in organisms. A TDRL consists of a tandem duplication of a contiguous set of genes followed by a random loss of one copy of each duplicated gene. Common sources of gene duplications include ectopic recombination, retrotransposition event, aneuploidy, polyploidy, and replication slippage.

123. Which one of the following combinations of terms is matched incorrectly?

(a) Nanopore : DNA sequencing

(b) Pyrosequencing : Protein primary structure

(d) SSRs : Co-dominant markers

Ans. (b)

Sol. Pyrosequencing is a method of DNA sequencing that detect light emitted during the sequential addition of nucleotides during the synthesis of a complementary strand of DNA.

124. Which one of the following statements is correct?

(a) If a transgenic plant heterozygous for an insert segregates into 1:1 ratio for the transgenic phenotype on back-crossing then it contains two unlinked copies of the insert.

(b) ANOVA allows a plant breeder to test whether measurements from three or more treatments show statistically significant differences.

(c) Comparative genomics allows scientists to identify regions of collinearity but not synteny between different species.

(d) For genetic mapping of a quantitative trait in plants, an RIL mapping population comprising of individuals that are heterozygous at most loci is preferred.

Ans. (b)

Sol. Analysis of Variance (ANOVA) test is a type of statistical test used to determine if there is a statistically significant difference between two or more categorial groups by testing for differences of means using variance. ANOVA tests are used when an individual wants to test more than two levels within an independent variable.

125. Which one of the following options represents a combination of terms that are matched INCORRECTLY?

(a) ddNTPs : Chain termination

(b) South Western blot: Physical interaction between DNA and proteins

(d) Yeast two hybrid system: Interaction between proteins

Ans. (c)

Sol. The 5' to 3' exonuclease activity can be coupled to the polymerization activity to displace DNA strands. However, the main function of the 5' to 3' exonuclease activity is to remove ribonucleotide primers that are used in DNA replication.

126. For a nuclear spin of spin quantum no. , precessing in a magnetic field at a Larmor frequency of 300 MHz, the wavelength of incident radiation required to excite the nuclear spins must be approximately.

(a) 1nm

(b) 1 cm

(d) 10 m

Ans. (c)

Sol. The correct option is (c).

127. Widal test, a widely used serological test for enteric fever, is a type of

(a) precipitation reaction

(b) agglutination reaction

(d) immunofluorescence detection

Ans. (b)

Sol. Widal Test is an agglutination test which detects the presence of serum agglutinins (H and O) in patients seru with typhoid and paratyphoid fever.

128. Fluorescence microscope that requires photoactivatable probes to obtain super-resolution is

(a) Structured illumination microscope (SIM)

(b) dSTORM – stochastic optical reconstruction microscopy

(d) Laser scanning confocal microscope

Ans. (b)

Sol. Single-molecule based localization microscopy methods, such as PALM, FPALM, STORM and dSTORM utilize photoactivatable or switchable fluorophores, which enable the temporal separation of fluoroescence emission.

129. A protein solution (0.2 ml) of unknown concentration was diluted with 0.8 ml of water. To 0.5 ml of this diluted solution 4.5 ml of biuret reagent was added and the color allowed to develop. The absorbance of this mixture taken in a test tube of 1cm diameter at 540 nm was observed to be 0.20.

0.5 ml of BSA (4 mg/ml) solution plus 4.5 ml of biuret gave an absorbance of 0.20 when measured as above. What is the protein concentration (mg/ml) in the undiluted unkown solution?

(a) 20

(b) 40

(d) 80

Ans. (a)

Sol. Initial diluted solution = 0.2 ml of unknown protein solution + 0.8 ml of water = 1.0 ml

0.5 ml of this diluted solution was added to 4.5 ml of biuret reagent (0.5 ml + 4.5 ml) = 5.0 ml

Given that, absorbance of unknown protein solution = 0.20

Again, solution of 0.5 ml of BSA solution + 4.5 ml of biuret = 5.0 ml

Absorbance of known BSA solution = 0.20

Given that, the concentration of known protein BSA = 4 mg/ml

Since absorbance are same for both solution. It means that 0.5 ml of unknown protein solution will also contain concentration of 4 mg/ml. Hence, 0.2 ml of unknown undiluted protein soluiton will also concentration of 4 mg/ml. So, 1 ml of undiluted solution will contain 4/0.2 = 20 mg/ml.

130. Given below are some terms and concepts related to phytoremediation in Column A and B

Which one of the following options represents the most appropriate match of all terms/concepts in column A and B ?

(a) A - II, B - IV, C - I, D – III

(b) A - III, B - IV, C - II, D – I

(d) A - IV, B - III, C - I, D - II

Ans. (d)

Sol. An excluder is a plant that has high levels of heavy metals in the roots but with shoot/root quotients less than 1. Plant metal chelators or phytochelatins (PCs) and metallothioneins (MTs) are the most common transporter proteins for heavy metal phytoremediation. Hyperaccumulators are unusual plants that accumulate particular metals or metalloids in their living tissues to levels that may be hundreds or thousands of times greater than is normal for most plants.

131. A culture was started by inoculating the medium with 100 cells having a generation time of 2 hours. Assuming the culture grows in ideal synchrony for at least 24 hours, what will be the number of cells in the culture at 2 hours and 9 hours?

(a) 2.0 × 10² cells, 1.6 × 10³ cells, respectively.

(b) 2.0 × 10² cells, 2.4 × 10³ cells, respectively.

(d) 2.0 × 10⁴ cells, 1.6 × 10⁸ cells, respectively.

Ans. (a)

Sol. As we know that,

N_t = N₀ × 2ⁿ; n is the number of generation

Number of cells in the culture at 2 hours = 100 × 2 = 200 = 2.0 × 10²

Number of cells in the culture at 9 hours = 100 × 2⁴ = 1600 = 1.6 × 10³

132. Pichia pastoris is a good host for producing human proteins for therapeutic use. Given below are some statements on the reasons for its utility.

A. It produces large amount of recombinant protein.

B. It has the property of secreting proteins into the medium.

C. It allows the formation of disulphide bonds similar to those in humans.

D. It carries out protein glycosylations identical to those found in humans.

Which one of the following options represents a combination of correct statements?

(a) A and B only

(b) A, B and C only

(d) B, C and D only

Ans. (b)

Sol. One of the main advantages of using P. pastoris as a protein production host is its ability to secrete high titres of properly folded, post-translationally processed and active recombinant proteins into the culture media. As a rule of thumb, proteins secreted in their native hosts will also be secreted in P. pastoris. Pichia pastoris produce large amount of recombinant protein. It has property of secreting proteins into medium and allows formation of disulfide bonds similar to those in humans.

133. The table below lists the biochemical characteristics of proteins and experimental procedures used to determine them. Match the characteristics with the experimental procedure.

Which one of the following matches is correct ?

(a) A - III, B - I, C - II, D – IV

(b) A - I, B - II, C - III, D – IV

(d) A - IV, B - II, C - I, D - III

Ans. (b)

Sol. NMR is used to determine the molecular structure at the atomic level of a sample. Apart from the molecular structure, NMR spectroscopy can determine phase changes, conformational and configurational alterations, solubility, and diffusion potential. IEF works by applying an electric field to protein within a pH gradient. The proteins separate as they migrate through the pH gradient in response to the applied voltage. When a protein reaches a pH value that matches its pI, its net electrical charge becomes neutral, and stops migrating. Affinity chromatography is a separation method based on a specific binding interaction between an immobilized ligand and its binding partner. Ultracentrifugation is a specialized technique used to spin samples at exceptionally high speeds. Current ultracentrifuges can spin to as much as 150 000 rotations per minute (rpm).

134. A circular dichroim spectrum in the far-UV region informs on the kind and content of secondary structures in a protein. Near-UV and tryptophan emission spectra inform on the tertiary structures. Shown in the panel above are (A) intrinsic fluorescence emission spectra of protein 'X', (B) Far-UV CD spectra of protein 'X', (C) Near-UV CD spectra of protein 'X' recorded under different conditions.

Curves represent the spectra of proein 'X' at pH 7.0 (black), pH 3.0 (green) and ph 7.0 in the presence of 6.0 M guanidine hydrochloride (red).

What does the experiment report?

(a) Protein is fully folded at pH 7.0, acid-induced molten globule at pH 3.0 and unfolded in 6M guanidine hydrochloride.

(b) Protein secondary structure is reduced at pH 7.0 and the protein has formed beta fibrils at the other two conditions.

(c) The changes in fluorescence and near-UV CD indicate increase in hydrodynamic radius at pH 3.0 and in 6M guanidine hydrochloride.

(d) There is extensive denaturation of the protein both at pH 3.0 and in 6M guanidine hydrochloride.

Ans. (a)

Sol. The correct option is (a).

135. The pI of four proteins (A, B, C, D) are shown in the table below:

To purify 'D' from a mixture of these four proteins in a single step, using ion-exchange chromatography, what combination of buffer pH and ion-exchange resin would you select?

(a) pH 11, cation exchanger resin

(b) pH 2, anion exchanger resin

(d) pH 8, cation exchanger resin

Ans. (d)

Sol. Protein A, B, C will have negative charges at pH = 8. Only protein D will have positive charge.

136. The following represents an equation for Bayesian stastics :

Which one of the following options correctly represents A, B, C and D in the above equation?

(a) A-Evidence, B-Posterior probability, C-Likelihood, D-Prior probability

(b) A- Likelihood, B-Prior probability, C-Posterior probability, D-Evidence

(d) A-Prior probability, B-Evidence, C-Posterior probability, D-Likelihood

Ans. (c)

Sol. The correct option is (c).

137. Given below are the various protein cleaving reagents (List I) and their recognition sites (List II) in the target protein.

Which one of the following options represents the correct combination of items

(a) A - III, B - II, C - I, D – IV

(b) A - IV, B - II, C - I, D – II

(d) A - III, B - IV, C - I, D - II

Ans. (d)

Sol. Chymotrypsin – Hydrolyzes the peptide bond at C terminal side of Phe, Tyr and Trp.

Cyanogen Bromide – Hydrolyzes the peptide bond at C terminal side of Met.

Trypsin – Hydrolyzes the peptide bond at C terminal side of lysine and arginines.

138. Given below are a few terms related to map-based sequencing of genomes:

A. Partial digestion with restriction enzymes.

B. Assembly of contigs

C. Pulsed field gel electrophoresis

D. Cloning in cosmids/YACs/BACs

E. Sub-cloning and sequencing

Which one of the following options represents the correct order of steps (based on the above terms) in map-based sequencing?

(a) C-B-D-E-A

(b) C-A-D-B-E

(d) A-C-B-D-E

Ans. (b)

Sol. Genetic mapping is based on the use of genetic techniques to construct maps showing the positions of genes and other sequence features on a genome. Genetic techniques include cross-breeding experiments or, in the case of humans, the examination of family histories (pedigrees). Option b is the correct order of steps in map-based sequencing.

139. Given below are a few statements on technologies/concepts related to development of transgenic plants:

A. Frequency of genetic transformation is influenced only by the genes of Agrobacterium and not by those of the host plants.

B. Transgenic plants containing a single copy of the transgene are preferred over those that contain multiple transgene copies for subsequent genetic analysis.

C. Supervirulent strains of Agrobacterium can be generated by increasing the copy number of virulence genes.

D. A nonconditional negative selection marker has to be necessarily used with a strong constitutive promoter for the development of transgenic plants.

Which one of the following options represents a combination of all INCORRECT statements?

(a) A and D only

(b) C only

(d) D only

Ans. (a)

Sol. A and D are incorrect statements.

140. Consider the four results that were obtained from immunophenotyping of human breast cancer cells.

A. B.

C. D.

Which one of the following options correctly depicts the above results?

(a) 'B' represents a plot that denotes a high percentage of cancer stem cells in the breast cancer cells.

(b) 'D' denotes a plot where dual positive cells predominate, representing the dead cells.

(d) 'C' represents a plot where only fibroblast cells are present.

Ans. (a)

Sol. CD44 expression is also upregulated in subpopulations of cancer cells and is recognized as a molecular marker for cancer stem cells (CSC). In humans, CD44 is encoded by 19 exons with 10 of these exons constant in all isoforms.

141. From the steps listed below, some are ued to evaluate the goodne of fit using the chi-square test.

A. The mean, variance and standard deviation are calculated

B. Variance calculated using

C. Test statitic calculated using

D. The degree of freedom is calculated as n–1, where n is the number of ways in which expected classes are free to vary

E. The probability value is obtained

Which one of the following options provides the correct sequence of steps in this stastical analysis?

(a) A, C, D

(b) C, D, E

(d) A, D, E

Ans. (b)

Sol. A chi-square test is a statistical test used to compare observed results with expected results. The purpose of this test is to determine if a difference between observed data and expected data is due to chance, or if it is due to a relationship between the variables you are studying.

142. Inverse PCR is performed for site-directed mutagenesis with complementary primers (having the desired mutation) using a plasmid having the cloned gene as template. The following statements were made regarding the above experiment.

A. PCR is followed by transformation of bacterial cells directly with the reaction mixture. A large number of the transformants will have the wild type gene.

B. The PCR mixture is treated with Dpn I and then used to transform bacterial cells. Most of the transformants will have the mutant gene.

C. PCR is followed by transformation of bacterial cells directly with the reaction mixture. None of the transformants will have the mutant gene.

D. The PCR mixture is treated with Dpn I and then used to transform bacterial cells. Half of the transformants will have the mutant gene.

Which one of the following options represents a combination of all correct statements?

(a) A and B

(b) A and D

(d) B only

Ans. (a)

Sol. Inverse polymerase chain reaction is a variant of the polymerase chain reaction that is used to amplify DNA with only one known sequence. Inverse PCR is a powerful tool for the rapid introduction of desired mutations at desired positions in a circular double-stranded DNA sequence.

143. The molecular ion peak [M]⁺ of an analyte as measure by Electron Ionization Mass Spectrometry has an m/z of 149 and a relative abundance of 100%. The [M]⁺ has a relative abundance of 6.7% and the [M + 2]⁺ peak has a relative abundance of 5%. The abundance of the major isotope of H, C, N, O and are ¹H-100%, ¹²C-98.9%, ¹³C-1.1%, ¹⁴N-99.6%, ¹⁵N-0.4%, ¹⁶O-99.8%, ¹⁸O-0.2%, ³²S-95.0%, ³³S-0.75% and ³⁴S-4.2%. The most probable molecular formula of the compound is :

(a) C₇H₁₂N₂O

(b) C₅H₁₁NO₂S

(d) C₆H₁₅NOS

Ans. (b)

Sol. The correct option is (b).

144. Which one of the statements given below regarding generation of monoclonal antibodies is INCORRECT?

(a) Monoclonal antibodies are the product of a single stimulated B-lymphocyte.

(b) To generate large quantity of monoclonal antibodies, a normal stimulated antibody producing B cell is fused with a long lived B cell tumor.

(d) For HAT selection of hybridoma, the antibody producing B-cells are pre-treated with 8-azaguanine to block salvage pathway of DNA synthesis.

Ans. (d)

Sol. Hybridoma technology is used to form hybrid cells from B-lymhocytes and tumor cells. After fusion, hybrid cells are selectively grown on HAT medium to separate from unfused cells. Unfused cells die on HAT medium and only hybridoma cells will survive. Aminopterin of HAT medium inhibits de novo pathway of nucleotide synthesis.

145. Optical remote sensing has been increasingly used to monitor vegetation globally. The table below lists different regions of the electromagnetic radiation (EMR) spectrum as well as different vegetation characteristics:

Which one of the following combinations correctly matche the EMR region of the vegetation character analysed :

(a) A - I, B - II, C - III,

(b) A - I, B - III, C – II

(d) B - III, C - II, D - I

Ans. (d)

Sol. Under high Photosynthetic active radiation, UV-B light increases the net plant photosynthesis in several plant species. The relationship between near-infrared reflectance at 800 nm (NIRR) from leaves and characteristics of leaf structure known to affect. All leaf water content indices utilizing strongly water absorbing wavelengths in the near-and shortwave-IR region.

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CSIR NET BIOLOGY (JUNE - 2021 Sift 1)
Previous Year Question Paper with Solution.